Suppose that \(\alpha _e=id_G\)  for all \(e\in E\)  and \(G\)  is right nilpotent. Then, \(E\)  is an ideal of \(B\)  and therefore by [1]} \(B^{(n)}=G^{(n)}+E\)  for all \(n\in \mathbb {N}\) . Therefore, if \(G\)  is right nilpotent and \(m\)  is a natural number such that \(G^{(m)}=0\) , we have that \(B^{(m)}=E\) .
